frac{1}{2 cdot 3} + frac{1}{4 cdot 5} + frac{ | vedclass

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(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)}$.Using partial fractions, $\frac{1}{(2n)(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1}$

Vedclass · Accepted Answer

$\log \left(\frac{e}{2}\right)$

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(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)}$.Using partial fractions, $\frac{1}{(2n)(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1}$.Thus, $S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{6} - \frac{1}{7}) + \dots$We know the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$For $x=1$, $\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots$Rearranging this, $1 - \ln(2) = 1 - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots) = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$Therefore, $S = 1 - \ln(2) = \ln(e) - \ln(2) = \ln\left(\frac{e}{2}\right)$.

$\frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \frac{1}{8 \cdot 9} + \dots$ is equal to

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